Hi, I’ve been stuck on this problem for a week and can’t seem to find any example that would help me to solve it. I would really appreciate it if someone would explain the steps to solving this problem. Thank you so much.

2. I developed a theorem (imcomplete) sometime ago, that if you have an integral linear combination of irrational distinct radicals equal to each other then there corresponding parts are equal.
For example, if
$1+\sqrt{2}+\sqrt{3}=x+y\sqrt{2}+z\sqrt{3}$
where $x,y,z$ are integers then,
$x=1,y=1,z=1$.
Now if you have,
$1+\sqrt{2}+\sqrt{3}=x+y\sqrt{5}$
where $x,y$ are integers is just impossible,
If you have that,
$\sqrt{16-6\sqrt{3}}=x+y\sqrt{3}$
where $x,y$ are integers then,
$16-6\sqrt{3}=(x+y\sqrt{3})^2$
Thus,
$16-6\sqrt{3}=(x^2+3y^2)+2xy\sqrt{3}$
Then,
$\left\{ \begin{array}{c}x^2+3y^2=16\\ 2xy=6$ is just impossible in integers.

This is my reasoning that why I would not expect a simpler solution with the integers.

3. Thanks for all your help and I'm am so sorry about breaking all those rules.

4. Originally Posted by coopsterdude
Hi, I’ve been stuck on this problem for a week and can’t seem to find any example that would help me to solve it. I would really appreciate it if someone would explain the steps to solving this problem. Thank you so much.

$
\sqrt{12-6\sqrt{3}}=3-\sqrt{3}
$

(using the convention that $\sqrt{}(\cdot)$ denotes the positive
root.)

Just goes to show that it never does to be too dogmatic PH

RonL

5. Originally Posted by CaptainBlack

Just goes to show that it never does to be too dogmatic PH
No I did not make a mistake in what I said.
Note I made a mistake in doing
$\sqrt{16-6\sqrt{3}}$ the wrong problem.
If you used my method then you will see that it works

-------------------------
I base it on the fact that,
$a_0+a_1\sqrt{b_1}+...+a_n\sqrt{b_n}$
is always irrational whenver all are integers and the b_i are all non-squares and a_i are not zero.

Then given
$a_0+a_1\sqrt{b_1}+...+a_n\sqrt{b_n}$= $c_0+c_1\sqrt{b_1}+...+c_n\sqrt{b_n}$
You can write,
$d_0+d_1\sqrt{b_1}+...+d_n\sqrt{b_n}=0$ where $d_i=a_i-c_i$
Which is rational an impossibility unless a_i=c_i giving zeros.

6. Originally Posted by CaptainBlack

$
\sqrt{12-6\sqrt{3}}=3-\sqrt{3}
$

(using the convention that $\sqrt{}(\cdot)$ denotes the positive
root.)

Just goes to show that it never does to be too dogmatic PH

RonL
I would really like to see the steps to get to the simplified form. Thank you for your time.

7. Originally Posted by Ranger SVO
I would really like to see the steps to get to the simplified form. Thank you for your time.
You can use my method which I shown, (I only did the wrong example which is why it did not work).
You attempt to find integers $x,y$ such as,
$\sqrt{12-6\sqrt{3}}=x+y\sqrt{3}$
Thus,
$12-6\sqrt{3}=(x+y\sqrt{3})^2$
Thus,
$12-6\sqrt{3}=(x^2+3y^2)+2xy\sqrt{3}$
Then, by my previous post,
$\left\{ \begin{array}{c}12=x^2+3y^2\\ -6=2xy$
Look at the second equation,
$-6=2xy$
This, is equivalent to
$xy=-3$
Since $x,y$ are integers the only possibilities are,
$(x,y)=(-1,3),(3,-1),(-3,1),(1,-3)$
Now we check which one of these satisfies the second equation,
we see that,
$\begin{array}{c} (3)^2+3(-1)^2=12\\ (-3)^2+3(1)^2=12$
Thus,
$3-\sqrt{3}$
$\sqrt{3}-3$
Both of these are true, indeed,
$(\sqrt{3}-3)^2=(3-\sqrt{3})^2=12-6\sqrt{3}$
But we need to chose,
$3-\sqrt{3}$
Because square roots are non-negative and,
$\sqrt{3}-3<0$

8. Thank you, I understand it now but I wonder if a High School Math Teacher would have intended a simpler solution.

I am finished with finals and had time to look at the problem.